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人気のある微分積分 >

integral of 5xsqrt(2x+3)

解

\int 5 x 2 x + 3 d x

解

(x - 1) (2 x + 3)^{\frac{3}{2}} + C

解答ステップ

\int 5 x 2 x + 3 d x

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 5 \cdot \int x 2 x + 3 d x

u置換積分法を適用する

= 5 \cdot \int \frac{u ( u - 3 )}{4} d u

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 5 \cdot \frac{1}{4} \cdot \int u (u - 3) d u

拡張

u (u - 3) : u^{\frac{3}{2}} - 3 u

= 5 \cdot \frac{1}{4} \cdot \int u^{\frac{3}{2}} - 3 u d u

総和規則を適用する:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 5 \cdot \frac{1}{4} (\int u^{\frac{3}{2}} d u - \int 3 u d u)

\int u^{\frac{3}{2}} d u = \frac{2}{5} u^{\frac{5}{2}}

\int 3 u d u = 2 u^{\frac{3}{2}}

= 5 \cdot \frac{1}{4} (\frac{2}{5} u^{\frac{5}{2}} - 2 u^{\frac{3}{2}})

代用を戻す

u = 2 x + 3

= 5 \cdot \frac{1}{4} (\frac{2}{5} (2 x + 3)^{\frac{5}{2}} - 2 (2 x + 3)^{\frac{3}{2}})

簡素化

5 \cdot \frac{1}{4} (\frac{2}{5} (2 x + 3)^{\frac{5}{2}} - 2 (2 x + 3)^{\frac{3}{2}}) : (x - 1) (2 x + 3)^{\frac{3}{2}}

= (x - 1) (2 x + 3)^{\frac{3}{2}}

定数を解答に追加する

= (x - 1) (2 x + 3)^{\frac{3}{2}} + C

グラフ

人気の例

integral of 1/(sqrt(x)(1+\sqrt{x))^3}

\int \frac{1}{x ( 1 + x ) ^{3}} d x

limit as x approaches 2 of ax^2-2bx+1

x \to 2 lim (a x^{2} - 2 b x + 1)

y^'-9/x y=((y^5))/(x^8)

y^{^{'}} - \frac{9}{x} y = \frac{( y ^{5} )}{x ^{8}}

derivative y=sin(5x)

d er i v a t i v e y = sin (5 x)

integral of 1/(2t)

\int \frac{1}{2 t} d t